The tangent to the curve γ€–y=eγ€—^2x at the point (0, 1) meets x-axis at:

(A)(0, 1)Β  Β  Β  Β  Β  (B) βˆ’ 1/2, 0

(C) (2, 0)Β  Β  Β  Β  Β (D) (0, 2)

The tangent to curve y = e2x at point (0, 1) meets x-axis at - Teachoo - NCERT Exemplar - MCQs

part 2 - Question 12 - NCERT Exemplar - MCQs - Serial order wise - Chapter 6 Class 12 Application of Derivatives
part 3 - Question 12 - NCERT Exemplar - MCQs - Serial order wise - Chapter 6 Class 12 Application of Derivatives

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Question 12 The tangent to the curve 〖𝑦=𝑒〗^2π‘₯ at the point (0, 1) meets x-axis at: (0, 1) (B) βˆ’ 1/2, 0 (C) (2, 0) (D) (0, 2) Given 〖𝑦=𝑒〗^2π‘₯ Finding slope of tangent at (0, 1) 〖𝑦=𝑒〗^2π‘₯ Differentiating w.r.t. π‘₯ π’…π’š/𝒅𝒙 = 2𝒆^πŸπ’™ At point (0, 1) Putting π‘₯=0, 𝑦=0 in 𝑑𝑦/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ = γ€–2𝑒〗^0 𝑑𝑦/𝑑π‘₯ = 2 Γ— 1 π’…π’š/𝒅𝒙=𝟐 Finding equation of tangent Equation of line at (π‘₯1 , 𝑦1) & having Slope m is π‘¦βˆ’π‘¦1=π‘š(π‘₯βˆ’π‘₯1) ∴ Equation of tangent at point (0, 1) & having slope 2 is (π’šβˆ’πŸ) = 2 (π’™βˆ’πŸŽ) (π‘¦βˆ’1) = 2π‘₯ π’š=πŸπ’™+𝟏 Now, we need to find the point where tangent meets the x-axis Since point is on the x-axis, it’s y-coordinate will be 0 ∴ Point = (𝒙,𝟎) Putting 𝑦 = 0 in (1) 0 = 2𝒙 + 1 βˆ’1 = 2π‘₯ (βˆ’1)/2=π‘₯ 𝒙 = (βˆ’πŸ)/𝟐 Thus, required point is ((βˆ’πŸ)/𝟐, 𝟎) So, the Correct answer is (B)

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo